Understand pointwise vs uniform convergence and their implications
A sequence of functions can converge one point at a time or all at once. The difference decides whether continuity and integrals survive the limit.
Pick a sequence, press Play to raise n, and move x. The N needed for a given epsilon changes from point to point.
Converges pointwise on [0,1] but not uniformly
For ε = 0.10, need N ≥ 4
A sequence of functions fₙ converges pointwise to f if for each x:
∀ε > 0, ∃N such that n ≥ N ⟹ |fₙ(x) - f(x)| < ε
Note: The N can depend on both ε and x. Different points may converge at different rates!
A sequence f_n converges pointwise to f if for each fixed x, the sequence of numbers f_n(x) converges to f(x). The rate can vary dramatically across different points.
Key insight: Pointwise convergence only requires convergence at each individual point — the speed of convergence can vary wildly from point to point.
Pick a sequence, turn on the ε-band, and raise n. Uniform convergence means the whole graph eventually fits inside the band.
sup|fₙ(x) - f(x)| does not → 0; no uniform N exists
fₙ → f uniformly if the same N works for all x simultaneously:
∀ε > 0, ∃N such that n ≥ N ⟹ sup|fₙ(x) - f(x)| < ε
Equivalently: ‖fₙ - f‖∞ → 0 as n → ∞.
Key insight: Uniform convergence means the entire graph of fₙ eventually fits inside an ε-tube around the limit function.
Uniform convergence requires a single N that works for all points simultaneously. The entire graph of f_n must eventually fit inside an ε-tube around the limit function.
Key insight: Uniform convergence means sup|f_n(x) - f(x)| → 0, controlling the worst-case error across the entire domain simultaneously.
Raise n with the limit function showing. Compare xn on [0,1], whose limit breaks apart, with sequences whose limits stay continuous.
Each fₙ is continuous, but the limit has a jump discontinuity at x=1. Pointwise convergence does NOT preserve continuity.
Theorem: If fₙ → f uniformly and each fₙ is continuous, then f is continuous.
The converse is false: pointwise convergence of continuous functions can yield a discontinuous limit (as seen with fₙ(x) = xⁿ).
Proof idea: Given ε, find N so ‖fₙ - f‖∞ < ε/3, then use continuity of fₙ.
Uniform convergence preserves continuity: if each f_n is continuous and f_n → f uniformly, then f is continuous. Pointwise convergence does NOT guarantee this.
Key insight: The uniform limit of continuous functions is continuous, but a pointwise limit of continuous functions can be discontinuous.
Pick a series and add terms. Each term stays under its bound Mn, and because Σ Mn converges, the partial sums settle uniformly.
Uniformly convergent since |cos(nx)/n²| ≤ 1/n² and Σ1/n² converges
Since Σ Mₙ converges, the M-test guarantees uniform convergence of Σ aₙ(x).
Theorem: Let Σ aₙ(x) be a series of functions on a domain D. If there exist constants Mₙ such that:
Then Σ aₙ(x) converges uniformly (and absolutely) on D.
Key idea: The Mₙ bound the functions uniformly, letting us control convergence independently of x.
A powerful tool for proving uniform convergence of series. If |a_n(x)| ≤ M_n for all x and Σ M_n converges, then Σ a_n(x) converges uniformly.
Key insight: The M-test reduces uniform convergence of a function series to convergence of a numeric series by bounding each term independently of x.
Pick an example and raise n, comparing the integral of fn with the integral of the limit. Sometimes they agree; sometimes they never do.
lim ∫fₙ dx
0.4000
∫(lim fₙ) dx
0.0000
Uniform convergence ⟹ we can interchange limit and integral. lim ∫fₙ = ∫f = 0.
Theorem: If fₙ → f uniformly on [a, b], then:
limn→∞ ∫ab fₙ(x) dx = ∫ab f(x) dx
Warning: Pointwise convergence is NOT enough! The counterexamples above show that ∫fₙ can stay constant even as fₙ → 0.
When can we swap the order of limit and integral? Uniform convergence guarantees lim ∫f_n = ∫ lim f_n, but pointwise convergence can fail spectacularly.
Key insight: Uniform convergence on a bounded interval lets you pass the limit under the integral sign, but mere pointwise convergence does not.
Differentiation interchange — requires f_n' → g uniformly and pointwise convergence at a single point to conclude (lim f_n)' = g
Spot a mistake or have an idea?